Question

A block is attached to one end of a spring with the other end of the spring fixed to a wall. The block is vibrating horizontally on a frictionless surface. If the mass of the block is 4.0 kg, the spring constant is k

Answer 1

The question revolves around a Physics concept from high school level concerning a block attached to a spring on a frictionless surface, exhibiting simple harmonic motion.

Step-by-step explanation:

The subject of the question is Physics, specifically related to simple harmonic motion (SHM) and mechanical systems involving a block and a spring. The student is inquiring about a system where a block of a certain mass is attached to a spring with a given spring constant, and how it vibrates when the opposite end of the spring is fixed to a wall. Such systems are often described using Hooke's Law, F = -kx, where F is the restoring force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position. An understanding of these concepts is generally taught at the high school level.

To solve problems like these, one must consider the potential and kinetic energy of the system at various points during the oscillation, the conservation of mechanical energy, and the equations governing SHM, such as x(t) = A cos(\(\omega t + \varphi\)), where A is the amplitude, \(\omega\) is the angular frequency, and \varphi\) is the phase constant.

Answer 2

Complete Question

A block is attached to one end of a spring with the other end of the spring fixed to a wall. The block is vibrating horizontally on a frictionless surface. If the mass of the block is 4.0 kg, the spring constant is k = 100 N/m, and the maximum distance of the block from the equilibrium position is 20 cm, what is the speed of the block at an instant when it is a distance of 16 cm from the equilibrium position?

Answer:

The velocity is
`v = 0.6 \ m/s`

Step-by-step explanation:

From the question we are told that

The mass of the block is m = 4.0 kg

The spring constant is k = 100 N/m

The maximum distance of the block from equilibrium position is d = 20 cm =0.20 m

The distance considered is
`d_k = 16 \ cm = 0.16 \ m`

Generally the maximum energy stored in the spring is mathematically represented as

`E = (1)/(2) * k * d^2`

=>
`E = (1)/(2) *100 * 0.2^2`

=>
`E = 2.0 \ J`

Gnerally according to the law of energy conservation

The energy maximum energy of the spring = energy of the spring at
`d_k` + energy of the block at
`d_k`

Here energy of the block at
`d_k` is mathematically represented as

`K_1 = (1)/(2) mv^2`

=>
`K_1 = (1)/(2) * 4* v^2`

=>
`K_1 = 2v^2`

Generally the energy of the spring at
`d_k` is mathematically represented as

`E_2 =(1)/(2) * k * d_k^2`

=>
`E_2 =(1)/(2) * 100 * (0.16)^2`

=>
`E_2 =1.28 \ J`

So

`2.0 = 1.28 + 2v^2`

=>
`v = 0.6 \ m/s`